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Sullivan  04 apcalc4e 45342 ch02 166 233 5pp  August 7, 2023  12:54



               170     Chapter 2 • The Derivative and Its Properties

                                                   Solution
                                                   (a) For c = 2,

                                                                                               2
                                                                        2
                                                                 f (x) = x − 5x  and     f (2) = 2 − 5 · 2 = −6
                                                      The rate of change of f at c = 2 is
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                                                                                      2
                                                                                                         2
                                                                    f (x) − f (2)   (x − 5x) − (−6)     x − 5x + 6
                                                          f (2) = lim          = lim               = lim
                                                           ′
                                                                x→2    x − 2    x→2      x − 2      x→2    x − 2
                                                                    (x − 2)(x − 3)
                                                              = lim             = lim (x − 3) = −1
                                                                x→2     x − 2     x→2
                                                                                    2
                                                   (b) If c is any real number, then f (c) = c − 5c, and the rate of change of f at c is
                                                                                          2
                                                                                 2
                                                                                                            2
                                                                                                        2
                                                               f (x) − f (c)   (x − 5x) − (c − 5c)    (x − c ) − 5(x − c)
                                                      ′
                                                     f (c) = lim         = lim                   = lim
                                                           x→c    x − c    x→c       x − c         x→c      x − c
                                                               (x − c)(x + c) − 5(x − c)  (x − c)(x + c − 5)
                                                         = lim                      = lim
                                                           x→c         x − c          x→c     x − c
                                                         = lim (x + c − 5) = 2c − 5
                                                           x→c
                                                                             R
                                                    NOW WORK   Problem 17 and AP Practice Problem 3.
                                                    EXAMPLE 3 Finding the Rate of Change in a Biology Experiment
                                                   In a metabolic experiment, the mass M of glucose decreases according to the function
                                                                              M(t) = 4.5 − 0.03t 2
                                                   where M is measured in grams (g) and t is the time in hours (h). Find the reaction
                                                   rate M (t) at t = 1 h.
                                                        ′
                                                   Solution
                                                   The reaction rate at t = 1 is M (1).
                                                                           ′
                                                                                          2
                                                                M(t) − M(1)     (4.5 − 0.03t ) − (4.5 − 0.03)
                                                       ′
                                                     M (1) = lim           = lim
                                                            t→1    t − 1     t→1          t − 1
                                                                      2
                                                                                          2
                                                                −0.03t + 0.03     (−0.03)(t − 1)    (−0.03)(t − 1)(t + 1)
                                                           = lim             = lim             = lim
                                                            t→1     t − 1      t→1    t − 1      t→1       t − 1
                                                           = −0.03 · 2 = −0.06
                                                      The reaction rate at t = 1 h is −0.06 g/h. That is, the mass M of glucose at t = 1 h is
                                                   changing at the rate of −0.06 g/h or decreasing at the rate of 0.06 g/h.
                                                    NOW WORK   Problem 43.

                                                    3 Find Average Velocity and Instantaneous Velocity

                                                   Average velocity is a physical example of an average rate of change. For example,
                             s  f (t)             consider an object moving along a horizontal line with the positive direction to the
                   0
                                                   right, or moving along a vertical line with the positive direction upward. The object’s
                        t
                             O                     location at time t = 0 is called its initial position. The initial position is usually marked
                                                   as the origin O on the line. See Figure 4. We assume the position s at time t of the
                            f (0)     s  f (t)  s
                                                   object from the origin is given by a function s = f (t). Here s is the signed, or directed,
                                                   distance (using some measure of distance such as centimeters, meters, feet, etc.) of
                                                   the object from O at time t (in seconds or hours). The function f is usually called
               Figure 4 t is the travel time. s is the
               signed distance of the object from the  the position function of the object. Motion along a line is sometimes referred to as
               origin at time t.                   rectilinear motion.

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